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# Hi guys, need some help on this calculus question :)

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Hi guys, I'm not too sure how to approach this question. Can anyone please give me some tips, etc.?

mathmeme  May 7, 2017
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### 4+0 Answers

#1
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$$y\cdot \arctan(e^{3x})=\dfrac{\arccos (x)}{x+2}$$ means $$y=\dfrac{\arccos (x)}{(x+2)(\arctan(e^{3x}))}$$

Then use implicit differentiation on the original equation:

$$y\cdot \arctan(e^{3x})=\dfrac{\arccos (x)}{x+2}\\ y'\cdot\arctan(e^{3x})+y\cdot\dfrac{1}{1+e^{6x}}\cdot (3e^{3x})=\dfrac{\left(-\frac{1}{\sqrt{1-x^2}}\right)(x+2)-\arccos x}{(x+2)^2}$$

Left hand side use product rule and chain rule and then right hand side use quotient rule(Wow I used every of them)

Gradient of curve y at point (x_1, y_1)= $$\dfrac{dy}{dx}|_{x=x_1}$$

You substitute $$y=\dfrac{\arccos (x)}{(x+2)(\arctan(e^{3x}))}$$ and then express y' in terms of x.

$$y'\cdot\arctan(e^{3x})+y\cdot\dfrac{1}{1+e^{6x}}\cdot (3e^{3x})=\dfrac{\left(-\frac{1}{\sqrt{1-x^2}}\right)(x+2)-\arccos x}{(x+2)^2}\\ y'\cdot\arctan(e^{3x})+\dfrac{\arccos(x)}{(x+2)(\arctan(e^{3x}))}\cdot\dfrac{1}{1+e^{6x}}\cdot (3e^{3x})=\dfrac{\left(-\frac{1}{\sqrt{1-x^2}}\right)(x+2)-\arccos x}{(x+2)^2}\\ y'\cdot\arctan(e^{3x})=\dfrac{\left(-\frac{1}{\sqrt{1-x^2}}\right)(x+2)-\arccos x}{(x+2)^2}-\dfrac{\arccos(x)}{(x+2)(\arctan(e^{3x}))}\cdot\dfrac{1}{1+e^{6x}}\cdot (3e^{3x})\\ y'=\dfrac{\dfrac{\left(-\frac{1}{\sqrt{1-x^2}}\right)(x+2)-\arccos x}{(x+2)^2}-\dfrac{\arccos(x)}{(x+2)(\arctan(e^{3x}))}\cdot\dfrac{1}{1+e^{6x}}\cdot (3e^{3x})}{\arctan(e^{3x})}$$

Then you substitute x = 0 for the last step. Do you want me to do it for you? :D

P.S. : I am an Asian

MaxWong  May 7, 2017
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You legend!

mathmeme  May 7, 2017
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OK nvm I do it for you XD

$$y'=\dfrac{\dfrac{\left(-\frac{1}{\sqrt{1-x^2}}\right)(x+2)-\arccos x}{(x+2)^2}-\dfrac{\arccos(x)}{(x+2)(\arctan(e^{3x}))}\cdot\dfrac{1}{1+e^{6x}}\cdot (3e^{3x})}{\arctan(e^{3x})}\\ \text{ when x = 0, }y'= \dfrac{\dfrac{\left(-\frac{1}{\sqrt{1-0^2}}\right)(0+2)-\arccos (0)}{(0+2)^2}-\dfrac{\arccos(0)}{(0+2)(\arctan(e^{3(0)}))}\cdot\dfrac{1}{1+e^{6(0)}}\cdot (3e^{3(0)})}{\arctan(e^{3(0)})}$$

The numerator equals -2 - pi/8 and the denominator equals pi/4 <-- that's basic trig and algebra

So the final answer is:

The gradient of the curve is $$-\dfrac{8}{\pi}-\dfrac{1}{2}$$ when x = 0

MaxWong  May 7, 2017
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Very impressive Max

Melody  May 7, 2017

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